Right then. I did some maths.
Now I’m assuming a perfect cone equates to one with a diameter (2r) equal to the height. That gives us an external surface area of
SA = pi * r * (sqrt(2r^2+2r^2))
Volume = 1/3 * pi * r^2 * 2r
According to ‘the internet’ the amount of energy falling on one m sq of the earth per second is 364J. Which gives us 3.14E+7 J per day (364*60*60*24). There are 343 days to xmas so that gives us a total energy of 1.08E+10 J falling on our idealised non cloudy average square meter of earth. However, it’s actually only falling for half that time (approx) on account of night. So we divide by two. 5.39E+9 J per year per m2.
Now the amount of energy required to raise 1g of water by 1 degree is 4.18J. Which is 4.18E+6 J to raise a m3 by 1 degree.
Having done some graphs it appears that it’s not just a mater of the size but also how cold it is to start with. If it’s minus 20 to start with, then it’ll always melt within the time frame stated.
To cut a long story short, it would appear that 220m in radius or 440m in diameter is the magic number assuming -20 to start with and that at 0 degrees it’s all melted. Excel can be supplied upon request!
Obviously this leaves out the density stuff.